From ebd7dd5bfbada4a66f18e27a18ae1fddb2daf0bc Mon Sep 17 00:00:00 2001 From: Jaron Kent-Dobias Date: Mon, 7 Dec 2020 14:15:51 +0100 Subject: Added lazy paragraph defending our choice of constraint by smearing the alternative. --- bezout.tex | 8 +++++++- 1 file changed, 7 insertions(+), 1 deletion(-) (limited to 'bezout.tex') diff --git a/bezout.tex b/bezout.tex index 02158a0..897e79d 100644 --- a/bezout.tex +++ b/bezout.tex @@ -31,7 +31,7 @@ \begin{equation} \label{eq:bare.hamiltonian} H_0 = \frac1{p!}\sum_{i_1\cdots i_p}^NJ_{i_1\cdots i_p}z_{i_1}\cdots z_{i_p}, \end{equation} -where the $z$ are constrained by $z\cdot z=N$ and $J$ is a symmetric tensor +where $z\in\mathbb C^N$ is constrained by $z^2=N$ and $J$ is a symmetric tensor whose elements are complex normal with $\langle|J|^2\rangle=p!/2N^{p-1}$ and $\langle J^2\rangle=\kappa\langle|J|^2\rangle$ for complex parameter $|\kappa|<1$. The constraint is enforced using the method of Lagrange @@ -41,6 +41,12 @@ multipliers: introducing the $\epsilon\in\mathbb C$, this gives \end{equation} At any critical point $\epsilon=H/N$, the average energy. +When compared with $z^*z=N$, the constraint $z^2=N$ may seem an unnatural +extension of the real $p$-spin spherical model. However, a model with this +nonholomorphic spherical constraint has a disturbing lack of critical points +nearly everywhere, since $0=\partial H/\partial z^*=-p\epsilon z$ is only +satisfied for $\epsilon=0$, as $z=0$ is forbidden by the constraint. + Since $H$ is holomorphic, a point is a critical point of its real part if and only if it is also a critical point of its imaginary part. The number of critical points of $H$ is therefore the number of critical points of -- cgit v1.2.3-70-g09d2